A proton and an electron have same de Broglie wavelength. Which of them moves faster and which possesses more kinetic energy ? Justify your answer.

Explanation

Let the subscripts 1 and 2 be used for proton and electron respectively.
`E_(1) = ( 1)/( 2) m_(1) v_(1)^(2) ` or `m_(1) v_(1)^(2) = 2E_(1)`
or `m_(1)^(2) v_(1)^(2) = 2E_(1)m_(1)` or `m_(1)v_(1) = sqrt(2 E_(1) m_(1))`
Now, `lambda = ( h)/(m_(1)v_(1)) = ( h )/(sqrt( 2E_(1) m_(1)))` or
`E_(1) = ( h^(2))/( 2m_(1) lambda^(2))`
`E_(2) = ( 1)/( 2) m_(2) v_(2)^(2) ` or `m_(2)v_(2)^(2) = 2E_(2)` or `m_(2)^(2) v_(2)^(2) = 2E_(2) m_(2)`
or `m_(2)v_(2)= sqrt(2E_(2)m_(2)) , lambda= ( h )/( m_(2)v_(2))= ( h )/( sqrt( 2E_(2) m_(2)))`
or `E_(2) = (h^(2))/( 2m_(2)lambda^(2)), `Dividing,`(E_(1))/( E_(2)) =( m_(2))/( m_(1))`
So, electron has more K.E. than the proton
Now, `( E_(1))/( E_(2)) = ((1)/(2) m_(1)v_(1)^(2))/( (1)/(2) m_(2)v_(2)^(2)) = ( m_(1)v_(1)^(2))/( m_(2)v_(2)^(2))` or
` ( v_(1))/( v_(2)) = sqrt((E_(1)m_(2))/( E_(2)m_(1)))`
But `E_(1) m_(2) lt E_(2) m_(1) :. v_(1) lt v_(2)` or `v_(2) gt v_(1)`
So , electron has greater velocity than the proton